We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
84
4
avatar

A spherical balloon is being inflated. Its volume is increasing at the rate of 2cm^3 per second. Find the rate in cm/sec at which the radius of the balloon is increasing when the radius of the balloon is 4cm. 

 

Any help would be appreciated :)

Thanks in advance,

Chelle

 Aug 3, 2019
 #1
avatar+103715 
+2

Hi Chelle,

A spherical balloon is being inflated. Its volume is increasing at the rate of 2cm^3 per second. Find the rate in cm/sec at which the radius of the balloon is increasing when the radius of the balloon is 4cm. 

 

\(V=\frac{4}{3}\pi r^3\\ \frac{dV}{dr}=4\pi r^2 \\~\\ \frac{dV}{dt}=2,\qquad \text{Find $\frac{dr}{dt}$ when r=4}\\~\\ \frac{dr}{dt}=\frac{dr}{dV}\cdot \frac{dV}{dt}=\frac{1}{4\pi r^2}\cdot 2=\frac{1}{2\pi r^2}\\~\\ so\\ \text{When r=4 the rate at which the radius of the balloon is increasing is }\quad \dfrac{1}{32\pi }\;\;cm/sec\)

.
 Aug 3, 2019
 #2
avatar
+1

Thankyou :)

Guest Aug 3, 2019
 #3
avatar+8521 
+1

A spherical balloon is being inflated. Its volume is increasing at the rate of 2cm^3 per second. Find the rate in cm/sec at which the radius of the balloon is increasing when the radius of the balloon is 4cm. 

 

\(r:\frac{4}{3}\pi r^3=\Delta r:\Delta V\\ 1:\frac{4}{3}\pi r^2=\Delta r:\Delta V\\ 1:(\frac{4}{3}\cdot \pi \cdot16cm^2)=\Delta r:(2cm^3/s)\\ \Delta r=\frac{2cm^3\cdot 3}{s\cdot4\pi \cdot 16cm^2} \) 

 

\(\Delta r=0,0298cm/s\)

laugh  !

 Aug 3, 2019
 #4
avatar
0

  Shouldn't second line be

4 pi r^2     (as the 3 brought down from the exponent cancels the 3 in the denominator)  ?   

Guest Aug 3, 2019

3 Online Users

avatar